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  • Applicants for the doctoral Programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections:

Applicants for the doctoral Programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths CM). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by ATE. AET is used by ATE for final selection.
For the 200 candidates who are at or above the 9oth Percentile overall based on CET, the following are known about their performance in CET:
 

1. No one is below the 80th percentile in all 3 sections.
2. 150 are at or above the 80th percentile in exactly two sections.
3. The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C The same is the number of candidates at or above the 80th percentile only in M.
4. Number of candidates below 80th percentile in V Number of candidates below 8oth percentile in C:
Number of candidates below 80th percentile in M = 4 : 2 : 1.
 

BIE uses a different Process for selection. If any candidate is appearing in the AET by ATE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by STE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P. 

Question : If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, what is the number of candidates who are at or above the 90th percentile overall and at or above the 80th percentile in both P and M in CET? 

Option: 1

60


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

200 candidates scored above the 90th percentile in CET.

The Venn diagram represents the number of candidates who scored above the 80th percentile in each of the three sections.

From (1), n = 0.

From (2), d + e + f = 150.

From (3), a = b = c.

Total candidates = 200.

Using these equations, we can deduce:

a + b + c + g = 50.

3a + g = 50, which implies a < 17.

From (4), (b + f + c) : (a + d + b) : (a + e + c) = 4 : 2 : 1.

Or, (2a + f) : (2a + d) : (2a + e) = 4 : 2 : 1.

6a + (d + e + f) = 7x.

6a + 150 = 7x.

a can be 3 or 10.

x can be 24 or 30.

2a + e can be 24 or 30, implying e can be 18 or 10.

2a + d can be 48 or 60, implying d can be 42 or 40.

2a + f can be 96 or 120, implying f can be 90 or 100.

3a + g = 50, so g can be 41 or 20.

Among candidates above the 90th percentile, those above the 80th percentile in at least two sections are selected for AET. Hence, candidates represented by d, e, f, and g are selected for AET.

BIE will consider candidates appearing for AET and above the 80th percentile in P. BIE will consider candidates represented by d, e, and g, which can be 104 or 80.

BIE will conduct a separate test for other students above the 80th percentile in P. Given that there are a total of 400 candidates above the 80th percentile in P, and since there are 104 or 80 candidates above the 80th percentile in P and at or above the 90th percentile overall, there must be 296 or 320 candidates above the 80th percentile in P who scored less than the 90th percentile overall.

From the given condition, g is a multiple of 5, so g = 20.

The number of candidates at or above the 90th percentile overall and at or above the 80th percentile in both P and M is equal to d + g, which is 60.

Hence, the correct answer is 60.

Posted by

Rishi

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